Mathematics: Evaluate the determinant for the following matrix:

To verify an identity means to prove that the equation is true by showing that both sides equal one another. identities are used in both course texts and in real life applications to abbreviate trigonometric expressions.Step-by-step explanation:Brainliest is appreciated...

Evaluate the determinant for the following matrix:

Ответ:

jahbreingram123

14.03.2022

Mathematics

ANSWER

The determinant is 0

EXPLANATION

For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth row:

$\det(A) = a_{k1} C_{k1} + a_{k2} C_{k2} + \ldots + a_{kn} C_{kn}$

where $a_{k1}$ is the entry of A in the kth row, 1st column, $a_{k2}$ is the entry of A in the kth row, 2nd column, etc., and $C_{kn}$ is the kn cofactor of A, defined as $C_{kn} = (-1)^{k+n} M_{kn}$ .

$M_{kn}$ is the kn minor, obtained by getting the determinant of the matrix which is the matrix A with row k and column n deleted.

Applying this here, we can expand along the 1st row. For convenience, let G be the matrix given by

$G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & 2 & 2 \\ 1 & 5 & 5 \end{bmatrix}$

where the first row has been bolded.

The determinant of G is therefore

$\begin{aligned} \text{det}(G) &= g_{11}C_{11} + g_{12}C_{12} + g_{13}C_{13} \end{aligned}$

Note that g₁₁ is the matrix element of G that is in the 1st row, 1st column,

g₁₂ is the matrix element of G that is in the 1st row, 2nd column, etc. Then we have

$\begin{aligned} \text{det}(G) &= g_{11}(-1)^{1+1}M_{11} + g_{12}(-1)^{1+2}M_{12} + g_{13}(-1)^{1+3}M_{13} \\ &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \end{aligned}$

M₁₁ is the determinant of the matrix that is matrix G with row 1 and column 1 removed. The bold entires are the row and the column we delete.

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ \bf 5 & 2 & 2 \\ \bf 1 & 5 & 5 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 2&2 \\ 5&5 \end{bmatrix} \right) \end{aligned}$

The determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

so it follows that

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ \bf 5 & 2 & 2 \\ \bf 1 & 5 & 5 \end{bmatrix} \implies M_{11} &= \det\left(\begin{bmatrix} 2&2 \\ 5&5 \end{bmatrix} \right) \\ &= (2)(5) - (2)(5) \\ &= 0 \end{aligned}$

Applying the same for M₁₂ and M₁₃, we have

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & \bf 2 & 2 \\ 1 & \bf 5 & 5 \end{bmatrix} \implies M_{12} &= \det\left(\begin{bmatrix} 5&2 \\ 1&5 \end{bmatrix} \right) \\ &= (5)(5) - (2)(1) \\ &= 23 \end{aligned}$

and

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & 2 & \bf 2 \\ 1 & 5 & \bf 5 \end{bmatrix} \implies M_{13} &= \det\left(\begin{bmatrix} 5&2 \\ 1&5 \end{bmatrix} \right) \\ &= (5)(5) - (2)(1) \\ &= 23 \end{aligned}$

so therefore

$\begin{aligned} \text{det}(G) &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \\ &= (1)(0) - (4)(23) + (4)(23) \\ &= 0 -4(23) + 4(23) \\ &= 0 \end{aligned}$

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Evaluate the determinant for the following matrix:

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